Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Organic Chemistry
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books & Services

Bookmark and Share
Lecture Comments (15)

1 answer

Last reply by: Professor Starkey
Sat Apr 25, 2015 8:44 PM

Post by Nikki Bell on April 25, 2015

Professor, at 7:40 should the nitrogen on the cyano group have a negative charge? After resonance the molecules still need to have a net neutral charge?

1 answer

Last reply by: Professor Starkey
Sun Aug 11, 2013 10:44 AM

Post by Jon Sorensen on August 7, 2013

maybe I'm just suffering from lack of sleep studying for my final but at 11:16 the 1,4 addtions of cuprates the final product looks 1,2 is the number based on nomenclature or the placements of substituents relative to one another?  again it maybe that I haven't slept for a long time sorry

1 answer

Last reply by: Professor Starkey
Thu Apr 18, 2013 11:15 AM

Post by Metavee Rynott on April 14, 2013

Hello I would like to know where the E+ from in 25:25. I do not see any of it in there. Thank you.

1 answer

Last reply by: Professor Starkey
Sat Apr 21, 2012 9:27 PM

Post by Kristen Medo on April 19, 2012

Please come teach at UT Austin!!

1 answer

Last reply by: Professor Starkey
Mon Jan 23, 2012 11:29 PM

Post by Jason Jarduck on January 22, 2012

Hi Dr. Starkey,

I was so excited with the first lecture I decided to watch the second lecture. Great lecture!!!

Thank You

Jason Jarduck

1 answer

Last reply by: Professor Starkey
Wed Nov 30, 2011 11:17 PM

Post by Gayk Gevorkyan on November 25, 2011

Professor, why doesn't the NU: attack the Carbonyl, but instead attacks the Beta-carbon?

0 answers

Post by Jamie Spritzer on August 16, 2011

in 22:46 it should be CH2 instead of CH

1 answer

Last reply by: Professor Starkey
Thu Oct 27, 2011 11:05 AM

Post by Senghuot Lim on July 24, 2011

Hey Dr Starkey, you're awesome and everyone should know it

Enols and Enolates, Part 2

Draw the product formed from this reaction:
Draw the product formed from this reaction:
  • This is a Michael reaction
Draw the product formed from this reaction:
  • This is a Ronbinson annulation reaction
Draw the product formed from this reaction:
  • This is a Ronbinson annulation reaction
Identify the starting materials for this Michael reaction:
Identify the starting materials for this Robinson annulation reaction:

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Enols and Enolates, Part 2

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Conjugate Additions 0:06
    • α, β-unsaturated Carbonyls
  • Conjugate Additions 1:50
    • '1,2-addition'
    • '1,-4-addition' or 'Conjugate Addition'
  • Conjugate Additions 4:53
    • Why can a Nu: Add to this Alkene?
    • Typical Alkene
    • α, β-unsaturated Alkene
  • Electrophilic Alkenes: Michael Acceptors 6:35
    • Other 'Electrophilic' Alkenes (Called 'Michael Acceptors)
  • 1,4-Addition of Cuprates (R2CuLi) 8:29
    • 1,4-Addition of Cuprates (R2CuLi)
  • 1,4-Addition of Cuprates (R2CuLi) 11:23
    • Use Cuprates in Synthesis
  • Preparation of Cuprates 12:25
    • Prepare Organocuprate From Organolithium
    • Cuprates Also Do SN2 with RX E+ (Not True for RMgX, RLi)
  • 1,4-Addition of Enolates: Michael Reaction 13:50
    • 1,4-Addition of Enolates: Michael Reaction
    • Mechanism
  • 1,4-Addition of Enolates: Michael Reaction 18:47
    • Example: 1,4-Addition of Enolates
  • 1,4-Addition of Enolates: Michael Reaction 21:02
    • Michael Reaction, Followed by Intramolecular Aldol
  • Mechanism of the Robinson Annulation 24:26
    • Mechanism of the Robinson Annulation
  • Enols and Enolates: Advanced Synthesis Topics 31:10
    • Stablized Enolates and the Decarboxylation Reaction
    • Mechanism: A Pericyclic Reaction
  • Enols and Enolates: Advanced Synthesis Topics 33:32
    • Example: Advance Synthesis
  • Enols and Enolates: Advanced Synthesis Topics 36:10
    • Common Reagents: Diethyl Malonate
    • Common Reagents: Ethyl Acetoacetate
  • Enols and Enolates: Advanced Synthesis Topics 38:06
    • Example: Transform
  • Advanced Synthesis Topics: Enamines 41:52
    • Enamines
  • Advanced Synthesis Topics: Enamines 43:06
    • Reaction with Ketone/Aldehyde
    • Example
  • Advanced Synthesis Topics: Enamines 45:31
    • Example: Use Enamines as Nu: (Like Enolate)
  • Advanced Synthesis Topics: Enamines 47:56
    • Example

Transcription: Enols and Enolates, Part 2

Hi, and welcome back to

Next, we are going to do Enols and Enolates, Part 2.0003

We are going to take a look at a reaction called a conjugate addition, and it involves an alpha, beta unsaturated carbonyl.0008

Now, so far we have been just studying carbonyls on their own (aldehydes, ketones, esters, and the reactions they undergo); but when we have a double bond conjugated to that carbonyl, it imparts some new reactivity.0015

Let's think about the resonance that this alpha, beta unsaturated carbonyl (also called an enone for short sometimes, because it has both the alkene and the ketone): what kind of resonance would that enone have?0026

Well, any carbonyl, we know, has resonance, so we know that this Lewis structure represents the structure and contributes to the overall hybrid of that enone.0039

But the positive charge, now, is an allylic one, so there is an additional resonance form we can draw--that resonance is extended through the conjugated π system, and we have a third resonance form.0054

So overall, when you look at the reactivity of an alpha, beta unsaturated carbonyl, we know that we have the usual partial plus, partial minus of a carbonyl; but what we find is that that partial plus character, that electron deficiency, is extended down into this β carbon, as well.0071

We have two electrophilic sites in an alpha, beta unsaturated carbonyl: that means, if a nucleophile is approaching this species, it has to decide between these two electrophilic sites on where to attack.0095

Now, certain nucleophiles (such as hydride, Grignard, organolithium...)--these are very reactive nucleophiles, and they prefer to attack the carbonyl as usual.0111

So, if they see an alpha, beta unsaturated carbonyl, these nucleophiles are going to totally ignore the double bond, the carbon-carbon double bond, and go after the carbonyl as usual.0122

We call this 1, 2 addition, because we have the nucleophile and the proton 1, 2 to each other (right next door).0134

In other words, they react like an ordinary ketone or aldehyde.0139

But, with the exception of hydride and Grignard, all other nucleophiles, when given an alpha, beta unsaturated carbonyl, will instead prefer to attack the β carbon.0144

We call this 1,4 addition, or conjugate addition, or Michael addition; we are going to see a reaction called a Michael reaction; sometimes this is called the Michael addition.0156

What that looks like is: the nucleophile doesn't attack the carbonyl; it attacks the β carbon; and it could do that because we can break this π bond and move it up, and then break this π bond and move it up.0166

So, our intermediate looks something like this: does this intermediate look familiar, when we have a double bond and an O- on the same carbon?0181

This is an enolate: we are going to get into enolate intermediate when we use these--when we do this conjugate addition.0194

And, when we do step 2 and we work this up, how would we protonate?0202

Well, we are not going to protonate up on the oxygen, because that would give us an unstable enol; instead, we are going to protonate...we are going to react just like an enolate does.0207

This lone pair is going to re-form the carbonyl, and it is the α carbon that is going to get protonated.0217

There is our product: that is what the product looks like for a 1, 4 addition.0233

If you take a look at protonating at the oxygen, just for comparison's sake, that would not be a stable product, because this would be an enol; so even if we tried to protonate in our workup on the oxygen, this would simply tautomerize back to the carbonyl.0237

OK, but this structure is nice, because maybe you can see why it is described as a 1, 4 addition: the nucleophile adds on the β carbon; the proton effectively adds on the oxygen; those are 1, 4 from one another.0258

So, that is why we call this 1, 4 addition versus 1, 2 addition, because all other nucleophiles other than hydride and Grignard prefer this 1, 4 addition, that is what we are going to be spending some time on.0271

OK, what nucleophiles?--things like enolates and also an organometallic reagent known as organocuprate.0285

Before we get into either of these examples, though, let's take a look at why it is OK for a nucleophile to add to this alkene.0294

We have never seen a nucleophile add to an alkene before; in fact, we have always seen alkenes as the nucleophiles themselves, and they go out and attack electrophiles.0301

And so, let's think about an ordinary alkene, and try and do this nucleophilic addition.0310

If a nucleophile tried to attack this alkene, we would get an intermediate like this; and that would not be a stable product--this is a very unstable C-.0315

There is nothing to help stabilize that negative charge, and so this would never happen: nucleophiles do not attack ordinary alkenes.0331

OK, however, if that alkene is part of an alpha, beta unsaturated carbonyl system, let's think about that reaction.0340

Same mechanism: let's just add to the alkene to get this carbanion.0346

And what is the difference now?--this carbanion is now allylic to the carbonyl, and is resonance-stabilized; we go back to recognizing the fact that we get this enolate intermediate--we get a resonance-stabilized intermediate.0354

That is why we call this thing an enolate.0375

OK, and that is why, in this case, nucleophilic addition is OK for an alpha, beta unsaturated alkene, but not for any other ordinary alkene.0379

This is really a reaction that is unique to enones.0391

We call enones Michael acceptors; and they are doesn't have to be a ketone that is attached to that carbonyl in order to allow for a 1, 4 conjugate addition.0397

There are other so-called electrophilic alkenes that would accept attack by a nucleophile; we could have (instead of a ketone) an any carbon-containing compound, or we could have a nitro or a cyano.0409

Now, we have seen the nitro, cyano, and carbonyl all being grouped together before: we call these EWGs, or electron withdrawing groups.0423

What they all have in common is: because of resonance, they all have partial positive character at this β carbon, so a nucleophile would be attracted to that site, and they all would stabilize the carbanion intermediate that would result from nucleophilic attack.0433

OK, what do they all have in common?--they all can withdraw electron density: the reason they are called electron withdrawing groups is that they withdraw electron density out of that double bond, putting a positive charge on that β carbon.0449

Any carbonyl can do it; any nitro can do it; so just looking at the resonance of the starting material, we can see right away why a nucleophile would be attracted to any of these species.0466

That is why we call them electrophilic alkenes, right?0483

So, that means this is an electrophile; this is electrophilic; this is electrophilic.0487

By touching any electron withdrawing group to a double bond, we make that double bond a Michael acceptor--something that is capable of undergoing 1,4 addition, conjugate addition.0499

Let's take a look at some of those reactions: one nucleophile that prefers to do this conjugate addition is called the organocuprate reagent.0511

Cuprates have the formula R2CuLi; so there are two carbon groups attached to a copper metal--that is what makes it an organic, organometallic reagent.0518

Let's see an example of such a reagent: this is called dimethyl cuprate, and like all organometallic reagents, it is a source of a nucleophilic carbon, as if that carbon (in this case, a methyl group) had a lone pair and a negative charge.0530

Now, because it is not ionic--it is actually coordinated to the metal--we put these around in quotes, to recognize that it is not ionic.0548

But it is useful to draw it as a methyl - , because that is very handy in our nucleophile, in our mechanism: it also reminds us that it is definitely a nucleophilic species.0561

We have a nucleophile in step 1: where do we have an electrophile?0571

Well, I know we always see a carbonyl and think it's an electrophile, but we need to expand our definition now and see that, when we have a conjugated, an alpha, beta unsaturated, carbonyl, that this is now going to be the electrophilic center of choice.0575

And so, what is going to happen is: my methyl - is going to attack here, break the π bond, and break the π bond.0592

Break this π bond...and so, it is going to give me an enolate intermediate.0606

And I form, in this case, a new bond between the β carbon and the organometallic carbon.0611

And so, in step 2, what is going to happen: we treat this with water; H3O+; we do our workup--how are you going to protonate this intermediate?0617

Well, it's an enolate, so it is the α carbon that is going to be reactive, so I'm going to use my O- to re-form the carbonyl and kick that π bond out to protonate the α carbon.0626

OK, and this is what we saw in the overview slide of what a conjugate addition looks like (a 1,4 addition): you break the π bond of the carbon-carbon double bond; you add the nucleophile to the β carbon; and you add a proton to the α carbon.0646

That is going to be the overall pattern for when we do a conjugate addition (1,4 addition).0667

So, cuprates allow us to add a variety of carbon groups to the β group.0672

Let's take a little closer look at cuprates: how do we make them, and what kind of behavior do we expect from them?0678

And also, what I want to point out is that the cuprates are nice to look at, because these organometallic reagents are in contrast with the other organometallic reagents we have studied up to this point.0685

The other organometallic reagents--something like a phenylmagnesium bromide...a Grignard--if we did a Grignard reaction, followed by H3O+, that, reacting with something like cyclohexenone, would prefer to do a 1,2 addition.0698

In other words, it attacks the carbonyl and gives an alcohol product.0712

OK, but instead, if I used a cuprate--if I used diphenyl cuprate (phenyl (2) CuLi), followed by workup, then we would expect it to add to the 1,4 position, because we have a partial positive here; we have a partial positive here; we have two electrophilic sites.0717

Cuprates are very useful, because they offer a complementary path to the other organometallic reagent--something that prefers to have 1,2 versus something that prefers to have 1,4.0734

OK, so where do cuprates come from?--well, they are actually prepared from organolithium reagents; organolithium reagents are prepared from alkyl halides.0746

OK, so if we wanted to make this dimethyl cuprate, we would start with methyl bromide or methyl iodide; we had added some lithium (remember, lithium prefers to do a halogen metal exchange).0755

So, we end up...that is how we form methyl lithium; and then, we take two equivalents of methyl lithium and treat it with some copper salts, and we get out this organocuprate structure.0766

We are always going to take two equivalents of the organolithium, because we need two equivalents of this alkyl group, or this carbon group, to be associated with the copper.0775

And one other interesting thing about the cuprates is that they can not only do a conjugate addition, but they will also do a coupling reaction with alkyl halides, or even aryl halides--some other carbon groups.0786

If you take this as your nucleophile, and you take a halide as your electrophile, you can get those two to couple and get an alkene product.0801

OK, this is not true for a Grignard reaction: Grignards don't do SN2 reactions, and this is not an SN2; this is a metal coupling reaction.0810

But this is very handy for synthesis; and again, it points to the difference in reactivities between a cuprate and an organolithium (or a Grignard reagent).0821

Now, the other nucleophile we would like to see added, besides just an ordinary carbon group, is an enolate nucleophile.0832

We call that reaction the Michael reaction, when we do a 1,4 addition, or a conjugate addition, where our nucleophile is an enolate and our electrophile is an enone.0837

Let's take a look at an example of one of these: if we took this diester and treated it with sodium hydride, sodium hydride is a good base, so that is going to deprotonate somewhere.0848

Where can we deprotonate?--there is only one place we could deprotonate, in this case--it is the α carbon.0864

So, let's just make that an anion there; that is what generates the nucleophile in the first step.0869

And, in our second step now, we are adding in our alpha, beta unsaturated ester, in this case; so that is our electrophile.0877

And so, let's see if we could draw a product.0886

Let's see if we could draw the product that we expect, and then we will take a look at the complete mechanism.0888

OK, we are going to form a new bond between this α carbon and this β carbon (it's the electrophilic carbon here).0893

Let's redraw our enolate: we have a new bond coming off from the α carbon, and then, what is it going to be attached to?--remember, what does it look like when a nucleophile does a conjugate addition to an enol?0903

You add the nucleophile to the β carbon, and then you add a proton to the α carbon.0917

And then, we have our methyl ester up here.0927

OK, so this is the electrophile you added to the α carbon; and you could also think that this is the nucleophile that you have added to the β carbon.0931

The Michael addition gets to be interesting--gets to be a little challenging--because you have both a complex nucleophile and a complex electrophile, and you need to understand the behavior of both in order to accurately draw the product.0944

OK, let's see what that mechanism looks like: our mechanism--we have our deprotonation of the α carbon; we used hydride, in this case, as our base (any base will do).0958

This is a one-way street, of course, because once we deprotonate...once you use hydride to deprotonate, you give off hydrogen gas, so this would be a way of 100% converting the diester to its enolate.0975

And then, in step 2, we bring in the electrophile; so we have our nucleophile; we have our electrophile; we have the α carbon attack the β carbon, break the π bond and move it up, break the π bond and move it up.0994

Eventually, you want that extra electron density to be up on oxygen; so that is why we keep moving our electrons up there.1015

We can redraw...we also have to decide, because they are both kind of complicated: we have to decide which one we keep right-side-up and which one we have to turn around.1022

And so, I always find it easier to redraw the nucleophile and the enolate standing upright; but then, we have to move this one down: so he's a CH2, single bond, CH, double bond, carbon.1031

And then, this carbon has an O- and an OCH3.1047

O-, OCH3...OK, so we just did our Michael addition; we just did our conjugate addition; now we need to protonate.1051

That was step 2, and now step 3: we add in a proton source; we do a mild aqueous workup, and we are going to protonate.1062

And here is where we have to be careful not to protonate on the oxygen; this is where doing the mechanism actually sometimes makes it a little harder than just trying to draw the product, because the mechanism might trip you up.1069

Remember, we are going to be protonating our α carbon.1081

We re-form our carbonyl, and it is the α carbon that gets protonated in our workup.1084

There is our new carbon-carbon bond; the β carbon, and now this was a CH; now, that is how it becomes a CH2, single bond, carbonyl, and then OCH3.1095

And let's see: does that match what we guessed in the first place?--it does; our enolate now has a new group attached to the α carbon, and our enone--in this case, it's an ester; our alpha, beta unsaturated ester now has a nucleophile added to the β carbon, and a proton added to the α carbon.1109

Let's see one more example: how about if I mix these two reagents in the presence of some base?1130

So, rather than doing step-wise, let's just throw it all in together; this is where, again, these problems can become challenging, because you have to decide what is going to happen first.1136

OK, clearly, I see that I have some base here, though--sodium methoxide is CH3O-, because we have an Na+.1145

So, I have a base; I need to go somewhere and deprotonate; where can I deprotonate?--I am going to find an α proton.1155

I am going to lose an α proton somewhere, and this is where it starts to get a little complicated, because you have an α proton here, and you have an α proton here.1165

Now, I don't see a carbonyl, but remember, this nitro as an electron withdrawing group is acting just like a carbon or even better, so this is an α proton just like this is.1175

OK, now how do I decide which one to go?--well, here is a hint I am going to give you: as soon as you see your alpha, beta unsaturated system here, you know that is going to be your electrophile.1188

OK, and you need to look elsewhere for a nucleophile to attack it; so it is going to be electrophile at the β carbon; I'm going to deprotonate over here at this α carbon; that is going to be my nucleophile; so now it is just a matter of drawing the product.1200

Without going through a whole mechanism, let's see if we can draw the product.1217

Let's start with our cyclohexenone, and if we want to do conjugate addition and Michael addition here, we break this π bond; the alkene carbon gets added to...we add our nucleophile to the β carbon, and what is the nucleophile that we are adding?1221

Well, we know we are going to deprotonate this CH, so now it's just a carbon, and what else does it have attached?--methyl, methyl, nitro.1238

CH3, CH3, NO2: and there is the product of our conjugate addition extending to something beyond just ketones, aldehydes, and esters.1247

Now, there is another reaction called the Robinson annulation, and this one involves two reactions in tandem.1263

It involves, first a Michael reaction (so a 1,4 addition, a conjugate addition), and then it's followed by an intramolecular aldol.1271

So, if you know what a Michael reaction looks like, and you know what an aldol reaction looks like, then you should be able to do a Robinson annulation.1278

The Robinson is always going to most often involves a substrate kind of like this: E is known as MVK--it needs some kind of MVK derivative (that stands for methyl vinyl ketone).1287

OK, here is a ketone: I have a methyl group on one side; I have a vinyl group on another side; and you will find that you have that component as part of your Robinson annulation.1306

There might be additional things attached to it, but you need to have at least both of those parts, and you will see why in just a moment.1314

OK, the first thing that happens: we are going to react this in some base (it can also be acid-catalyzed, but we will take a look at the base one here); the first thing that happens is a Michael reaction.1321

A Michael reaction means I have a conjugate addition of an enolate; so like I said, every time we see an enone, you know that is going to be your electrophile, or MVK is our electrophile here, so where is a nucleophile?1331

Right over here, we have our nucleophile, our α carbon.1345

The first thing that is going to happen is (let's bring them down here) a Michael, which means we are going to have a new group to this α carbon.1348

We are going to have a CH2 and a CH, and we are going to draw it around this way.1362

We are going to flip it over this way to keep it positioned as it was right here, because now this is perfectly situated; we are not going to stop here, because this intermediate structure is perfectly situated to do an aldol reaction.1371

What does an aldol mean?--it means we have an enolate, the α carbon on one attacking the carbonyl carbon on the other.1388

We have an α carbon right here; there is our nucleophile; and 1, 2, 3, 4, 5, 6 atoms away, very conveniently, we have an electrophile.1395

It will always end up 6 atoms away if we have a methanol/ketone sort of electrophile that we are adding to.1408

What happens is: we now have an aldol reaction; we form a 6-membered ring; and it almost always involves the loss of water.1415

When we have a cyclic structure like this, the loss of water is even more favorable.1427

And so, almost always, we get this second reaction happening with loss of water.1433

And so, the pattern we get in our product is a cyclohexenone product.1441

The Robinson annulation will always be forming a 6-membered ring; now notice, we already started with a 6-membered ring, so that is why we have a second one now.1451

We form a 6-membered ring, and it is always a cyclohexenone; it has the double bond and the carbonyl.1458

Let's see if we can do a mechanism for this: we are in base, so our first step is going to be deprotonate; we need to get started by doing the aldol reaction.1467

I am going to deprotonate my α carbon, using methoxide as my base.1478

So, step 1 is: Form an enolate.1486

An enolate, of course, is a nucleophile, so we look around for an electrophile.1497

The methyl vinyl ketone is the other ingredient we had in our reaction, so that is our electrophile.1502

What does the nucleophile do?--α carbon to this β carbon: that is the bond we are forming.1509

Enolate attacks, breaks the π bond, breaks the π bond...look at that: 4 arrows all on a single step of the mechanism.1517

This is the case; now again, when it comes to having errors, this is a very easy place to lose carbons; so let's check: we have a 1, 2, 3, 4-carbon chain (right?--methyl vinyl ketone).1528

So, let's make sure we are adding 1, single bond, 2, double bond, 3, O-, single bond, 4.1543

You still need to have a 4-carbon chain that you are adding on top of the six carbons we already had.1555

This is really the #1 most common place that we can accidentally lose carbons with our line drawings.1562

There is our conjugate addition; there is our Michael addition; and now, we need to protonate to finish up this mechanism (this part of the mechanism).1569

Let's bring in some methanol to protonate our α carbon, and we just finished doing an aldol.1577

So, what were the three steps in the aldol mechanism, the base-catalyzed?--deprotonate, and then attack, and then reprotonate.1593

Deprotonate, attack, reprotonate: we see that pattern so many times. 1606

OK, we did our aldol reaction; this was all an aldol; OK, but we don't stop here, because our product is perfectly set up to continue doing another attack.1611

And here is where we need to recognize: the new bond that is going to be formed is with the methyl group of the methyl vinyl ketone.1624

That is 1, 2, 3, 4, 5, 6 atoms away from the carbonyl of our initial enolate.1632

This is going to be our nucleophile #2, if you will; electrophile #2; and that is the bond we need to form.1640

So, how does that happen?1652

I'm sorry--excuse me: I just realized we are about to do an aldol; that is because we just did a Michael addition; we did the conjugate addition of the alpha, beta unsaturated carbonyl.1654

It starts with a Michael, and now we do an aldol.1664

So now, we have the α carbon attacking a carbonyl; so how do we get started?1667

Well, we get started the usual way: we deprotonate; let's bring our base in again...methoxide...we deprotonate to make an enolate.1671

And now, if you would like, you can flip your enolate around so it's a little easier to see the 6-membered ring you are about to form.1682

Here are our 1, 2, 3, 4 carbons still; now we can see our nucleophile and our electrophile a little more clearly.1690

The bond that you are about to form is right there.1697

We deprotonate, and now we attack (enolate attacks the carbonyl--that is the classic aldol), and now we just formed a new 6-membered ring.1703

This is a double bond up here, a carbonyl; these are the two carbons that were just involved; and what does this carbon have on it?--single bond, O-.1723

What do we have to do to finish up?--all we need to do is protonate.1737

I'm going to use our methanol again; and we are done.1745

Well, we are almost done, I should say.1755

OK, we did our Michael; we did our aldol; and what is left?--well, this is not the typical product of Robinson annulation; this a β-hydroxy carbonyl, but typically, we get the cyclohexenone product out, so we have to lose water.1758

Let me redraw this down here so we have some space for that mechanism.1777

The last thing we want to do is: we want to dehydrate.1786

OK, let's think back to that aldol: when we wanted to dehydrate, how many steps was that?1794

That was a 2-step mechanism; we call that β elimination.1798

And how did it start?--it starts like all of these base-catalyzed ones do.1804

We are going to deprotonate an α carbon; if you recognize the similarities to the patterns, that is really going to help.1809

Deprotonate to form an enolate, and then (so we are going to say step 1 is deprotonate) what does this enolate do?1816

It sees that β leaving group, so it re-forms a carbonyl, kicks down the π bond, and kicks off that hydroxide leaving group.1834

Step 2 is "eject the β leaving group."1843

Now, we have our aldol, our Robinson product; so Michael first, and then we did an aldol with dehydration second--quite a long mechanism, but if you can do the base-catalyzed Robinson annulation mechanism, that means you have really mastered all the various reactions that enols and enolates can undergo.1849

Now, let's talk about some advanced synthetic techniques that involve enols and enolates.1871

There is one that involves stabilized enolates, and a reaction known as a decarboxylation reaction.1877

OK, so let's do this decarboxylation reaction first.1885

If you have a β-keto acid (so in other words, a carboxylic acid that--not in the α position, but in the β position, you have a carbonyl, you call that a β-keto acid for short), it turns out that, when you heat these, they lose CO2.1888

You take this carbon dioxide group; this gets lost as CO2; this one carbon and two oxygens gets lost as CO2.1902

And so, your product--it is as if you had just erased that functional group from there.1911

The α carbon is still intact, but the extra carbon is gone.1917

It is known as the decarboxylation reaction: this could be very useful to us, synthetically; and we will see examples of that.1921

Now, how does this happen?--the mechanism is a single step concerted mechanism; it's an example of a pericyclic reaction.1929

If we redraw our carboxylic acid like this, we can see a favorable 6-membered transition state here that can be formed.1937

This carbonyl could come up and grab that proton, which can break that OH bond and form a π bond and break this carbon-carbon bond.1946

So ultimately, that is what you want to do: you want to break that carbon-carbon bond until you have cleared off the molecule of carbon dioxide.1954

So, when we follow these arrows around, we end up with this product as one product and this product as another product.1961

That is the result of that pericyclic step; and now let's think about what these products are.1973

This one is just carbon dioxide (drawn a little strangely, but that is carbon dioxide); so we ejected that molecule, and what about this structure--how would you describe this functional group?1977

It has an alcohol on the same carbon as a double bond; this is an enol, and what do we know enols do?--they tautomerize very favorably to the corresponding ketone, which was the ketone that we were expecting in this reaction.1988

OK, so the mechanism is a loss of CO2 followed by our two-step tautomerization mechanism; that is the complete mechanism.2003

Now, let me show you a reaction sequence where this is going to be useful to us: we can start with a substrate like this, called diethyl malony.2014

If you treat this with a two-step process (sodium ethoxide and benzobromide), the first step: you deprotonate to make the enolate.2022

OK, now remember, this is a stabilized enolate, because it has 2 electron withdrawing groups; we could use a very weak base--mild base, like ethoxide--to completely 100% deprotonate it, and that reaction goes well.2034

This is very stable, and that means it is not only easy to make; it is also easy to use, and this alkylation reaction, then, is typically going to be pretty high-yielding.2048

So, we know that our enolate is a nucleophile, so when we add in our alcohol electrophile, we can get an SN2; we can do our alkylation reaction here.2060

OK, and we have seen this before, doing an α alkylation; but here is the interesting part: if you take this species, now, and treat it with H3O plus heat, the H3O+ first does a hydrolysis; it takes our ester groups and converts them to carboxylic acids.2077

So, we have addition-elimination of the water; we get a substitution.2099

We get this diacid out, and what happens when we heat this?--well, we have our β carbonyl carboxylic acid pattern again.2106

We have a carboxylic acid, and not in the α position, but in the β position, we have a carbonyl.2120

Now, in this case, it is not a ketone--it is another carboxylic acid--but that is OK; any carbonyl will do.2125

And then, what happens is: we eliminate one of these carboxylic acids (it's symmetrical--it doesn't matter which one); we eliminate one of them as carbon dioxide.2131

Here is our α carbon; our α carbon still stays intact, but the CO2 group that used to be attached with the carboxylic acid group is now gone.2147

This overall sequence is pretty high-yielding and goes pretty well; and so, it turns out that this is a very useful strategy for synthesizing target molecules that look like this, that might otherwise be difficult to achieve.2156

This diethyl malony acts as if, instead know that one of these carbons can be lost as CO2; so it reacts just like you had this enolate.2171

If you had the acetic acid enolate, then you could make some very interesting molecules; well, the diethyl malony kind of gives you that structure, because it gives you the two carbons with the α carbon having a negative charge.2193

This extra carbon--we add in that extra carbon to help stabilize this enolate.2210

It is added as an activating group that can later be removed; OK, and of course, the ethoxy group is like a protector group for that OH and makes it more stable.2216

This is our acetic acid enolate equivalent; we describe these two as being synthetic equivalents.2227

Any time you want to use the acetic acid enolate, we use diethyl malony instead.2242

OK, ethyl acetoacetate does the same thing; remember, this group can be clipped off and lost as carbon dioxide; so this is a very nice reagent to use to be the acetone enolate equivalent, because the enolate of acetone is not so easy to use.2246

It's very reactive--it's a pretty strong base--and so, the yields in its reactions are not very high.2269

So instead, what we do is: we add an ester group on here to act as an activating group to help stabilize the enolate intermediate.2274

And then, we have higher-yielding reactions, and we can remove that group when we are done.2281

Let's look at an example where this might be useful; I think I want to do the following transform.2288

Let's make this carboxylic acid from this bromide.2301

OK, so it looks like we comparing our starting material and our product, we see that this is a new carbon-carbon bond that we need to form, because we just have these five carbons initially.2311

So, if I wanted to get these two carbons to come together, one of them must have been a nucleophile; one of them must have been an electrophile.2323

This one is α to a carbonyl, so that makes it a very reasonable nucleophile; so let's assume that this was my nucleophile, since it was an α carbon, which means this was my electrophile.2334

And so, if we do a retrosynthesis, our retrosynthesis is asking, "What starting materials I need?"--we need to plan the synthesis before we attempt doing this transform.2349

What we can do is: we can go right back to just synthons; let's just break that bond and put the positive charge here and the negative charge here.2362

These are called synthons; and if you had these imaginary cation and anion, they would obviously come together to make this product.2374

Well, so we can't buy the cyclopentyl carbocation, but that is the synthetic equivalent of having, let's say, a good leaving group on that carbon, which happens to be exactly what we have.2382

So, that would be a good electrophile that we already have.2395

And to make this nucleophile, what we are going to use, instead of trying to deprotonate acetic acid (which would be impossible): we are going to use diethyl malony.2400

OK, and so, what the synthesis looks like, then, is: we are going to use the anion, the enolate, of diethyl malony; so we'll treat this first, let's say, with some sodium methoxide, to make the anion; and then, we will throw in the alkyl halide.2416

OK, and this can do an SN2; you expect that reaction to happen pretty well, with an ordinary enolate that would...because this would do an elimination; but because this is more stable and less reactive, we can do this SN2.2431

OK, but how is this related to this compound?--well, we still see our two carbons (our 1,2 carbons); all we have to do is get rid of that extra carbon; so all we do is H3O+ and heat; that is going to hydrolize, and that is going to lose carbon dioxide.2453

Now, in practical measures, it turns out that this saponification reaction of a hydrolysis of an ester usually works better with sodium hydroxide (with base).2472

So probably, what we would do is sodium hydroxide and water to do the hydrolysis more easily.2482

And then, step 2 is: we can reprotonate to get the diacid, and it is the diacid that, upon heat, will do the decarboxylation reaction.2489

These are some advanced synthetic techniques involving the enolates, where we can use synthetic equivalents to make our enolates easier to form and easier to handle--higher-yielding reactions.2497

Now, there is another advanced synthetic topic, and that is dealing with enamines; this is another functional group that is related to the enols and enolates, so this would be a good time to talk about it.2513

It turns out, these are also synthetic equivalents of enolate nucleophiles: so let's think about where an enamine comes from.2528

First, let's recall: if you have a ketone or an aldehyde, and you react with a primary amine or ammonia and some kind of acid, you replace the C-O double bond with a C-N double bond.2534

We call this imines, when we have a C-N double bond and the reaction is an attack of the nitrogen nucleophile on the carbonyl, and then collapse of CTI to kick that water out.2548

So overall, you are losing water in this reaction.2561

In an imine, when you have a C-N double bond and you treat it with H3+ hydrolysis, this will revert back to the ketone and aldehyde, give me back the carbonyl, and free up that amino group.2566

OK, so we have seen this reaction before, of a primary amine with a ketone or an aldehyde.2579

Now, instead of a primary amine, if you use a secondary amine, then that is going to give a different product.2587

When you erect it with a ketone, you are going to again replace the C-O double bond with a C-N double bond; but this product is now not going to be stable, because that nitrogen has four bonds; it's going to be positively charged.2594

So, how can you stabilize this?--well, you can deprotonate one of these α protons, and that will get rid of the positive charge on nitrogen and give you a structure something like this.2606

OK, this is known as an enamine, because it has an alkene and carbon-carbon double bond and an amine on it.2621

It kind of looks a little bit like an enol, right?--instead of having an oxygen here, we have a nitrogen; OK, but it is definitely related to that.2630

It is going to be related like an enol and an enolate, in being nucleophilic.2641

So first, let's just do an example of this: if we reacted this ketone with a secondary amine (it's acid; this is tosic acid; it's just...any old acid could be used here)--OK, what is going to happen is: we are going to replace the C-O double bond with a C-N double bond.2648

But that is not going to be our final product: we can't stop there--we need to neutralize this.2669

But this is a case where we should think about where to deprotonate--which of these α protons would be better.2674

What we are going to do is: we are going to want to go for this proton, because that is going to give the more stable enamine product out.2684

And so, this is a feature of enamine chemistry: you get the more stable structure out (known as the thermodynamic enol equivalent).2694

OK, so why would having a double bond here be more stable--be better than having a double bond here?--well, because it's conjugated now with the benzene ring: the benzene ring helps conjugate that, so we can conjugate the double bond to make it more stable.2707

We can have more α groups on there: more highly substituted makes it more stable.2720

Conjugated π bond would be the major product here.2725

Now, what can we do with an enamine?--we are going to react it just like an enolate; so if we took this enamine that we just formed, this would be a nucleophile.2733

So, if we reacted it with a reactive electrophile (like an alkyl halide, primary alkyl halide, or a methyl halide, for example), they are going to react, and where is an enolate or an enol reactive?2742

It is at the α carbon, so right there is where it is going to be nucleophilic.2757

How do we show that react?--well, analogous to an enol or an enolate, it is the lone pair that is going to come and form a π bond; that is going to kick this π bond out and attack whatever electrophile it is that we are going after (in this case, it's an alkyl group).2763

We are doing an α alkylation.2780

Oh, I'm sorry, we lost our π bonds here; this was a benzene ring; we get the same product either way.2785

This would give us the following intermediate, after we did this α alkylation.2797

OK, and this would not be stable; we would need to do something to stabilize this, and typically what we are going to do: if our methyl iodide was step 1, step 2--we are going to treat this with H3O+.2802

So, rather than just neutralize it, we are going to do a hydrolysis.2815

That takes the C-N double bond and turns it back into a C-O double bond.2821

Ultimately...if you remember, we started with this ketone; and so, the transformation we did was to go from this ketone and alkylate at this position.2834

Perhaps we could have done that using LDA, but depending on the ketone, depending on the substrates you are using, sometimes it is preferable instead to use the enamine.2847

Especially when you are trying to control regiochemistry, you could use the enamine as your nucleophile, instead.2857

In fact, enamines predated LDA; so this is the strategy that we could have had/could have used before LDA was an option for us to deprotonate and form the enolate.2864

Let's see if we can do one more example here: we can take cyclopentanone and react it with this amine.2879

A secondary after step 1, we are going to convert it to an enamine; so the nitrogen takes the place of the oxygen, and we get a double bond.2892

In this case, it doesn't matter which side it goes to.2904

This is an enamine; this is a nucleophile; and so, hopefully, we are looking for some kind of electrophile now.2909

Our electrophile and our second step is benzoyl chloride.2917

How can we bring those two together?--well, here is our α carbon; so we can have our lone pair come down and our π bond come out.2928

And then, do we lose the leaving group?2939

No, we don't lose the leaving group: we break the π bond and go up here; remember, every time we attack a carbonyl, we break the π bond.2942

OK, that kind of brings us to this complex structure here, a little bit.2959

We have an acid chloride that we are reacting with, so what happens when we break that π bond to the carbonyl?2963

This intermediate is not very stable: this is most definitely a CTI, a charged tetrahedral intermediate, with a very good leaving group attached to the O-.2971

So, for sure, this oxygen is going to come back down, re-form the carbonyl, and kick out that leaving group.2979

Always, with an acid chloride, we do addition-elimination; we do substitution for that excellent leaving group.2987

That would be a way of installing this acyl group to the α carbon; that is pretty handy.2997

And then, our last step here: we are going to do hydrolysis.3004

An enamine is something that we create temporarily to use and activate it and make it like an enolate.3008

And then, we take that nitrogen back out.3017

That is what makes it a synthetic equivalent.3020

Step 3: we are going to do hydrolysis; and hydrolysis of an imine gives us back our carbonyl; and that would be a very nice way of doing this transformation.3023

Instead of using LDA and a very reactive enolate, we can use an enamine instead and react it with an acid chloride (electrophile, in this case).3038

That wraps it up for our second part of Enols and Enolates.3049

Thanks very much for visiting

We hope to see you again soon; thank you.3054